SOLUTION: find the area of a triangle with the vertices ( 7.-9), (-1,6),(-1,-9)

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Question 1045419: find the area of a triangle with the vertices ( 7.-9), (-1,6),(-1,-9)
Found 3 solutions by josgarithmetic, advanced_Learner, MathTherapy:
Answer by josgarithmetic(39616) About Me  (Show Source):
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Use the length from (7,-9) to (-1,-9) as a base of the triangle.
Use the length from (-1,6) to (-1,-9) as the height of the triangle.

If b is BASE and h is HEIGHT, then formula for area of a triangle is %281%2F2%29bh.

Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
let A( 7.-9), B(-1,6),C(-1,-9)
AB=sqrt%2864%2B225%29
AB=%2817%29
BC=sqrt%280%2B225%29
BC=%2815%29
AC=sqrt%2864%2B0%29
AC=%288%29
Since8%5E2%2B15%5E2=17%5E2
the hypotenus is 17
Area=1%2F2%2A8%2A15
area=60

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

find the area of a triangle with the vertices ( 7.-9), (-1,6),(-1,-9)
Let points (7, - 9), (- 1, 6), and (- 1, - 9) be A, B, and C, respectively
We then have: A(7, - 9), B(- 1, 6), C(- 1, - 9)
Line AC is parallel to the x-axis, and so, its length is: 7 - - 1, or 7 + 1, or 8 units
Line BC is parallel to the y-axis, and so, its length is: 6 - - 9, or 6 + 9, or 15 units
We now see that the polygon is a right-triangle, and that AC and BC are its legs.
Therefore, area = %281%2F2%29+%2A+8+%2A+15, or highlight_green%28matrix%281%2C3%2C+60%2C+sq%2C+units%29%29